Problem: Rachel, being an awesome android programmer, just finished an App that will let us draw a triangle by selecting three points on the touch-plane.Now She asks her friend Bruce to draw a Right-Angled Triangle (we'll call it RAT) by selecting 3 integral points on the plane.
 A RAT is a triangle with Non-Zero area and a right-angle.

But Bruce, being in a hurry as always, makes a mistake on drawing the points. Now Rachel's app can fix Bruce's mistake if only one coordinate of any one point can be adjusted within a distance of 1 units. i.e., if the drawn triangle can be made a RAT by adjusting any one coordinate within 1 unit of distance.

Your task is simple, given 6 coordinates (each Integer pair representing a point), you have to tell can Rachel's app correct Bruce's mistake automatically?

Input:

      First line containing T, the number of test cases.
      Then T lines follow with 6 space separated coordinates.

Output:

      print "YES" in a newline if it's possible.
      otherwise print "NO" in a newline.

Constraints:

      1 ≤ T ≤ 100
      -1000 ≤ xi,yi ≤ 1000

SAMPLE INPUT
3
-1 0 2 0 0 1
2 3 4 5 6 6
0 0 2 0 10 10

SAMPLE OUTPUT
YES
NO
NO

Explanation

Case 1:  For (-1, 0), (2, 0) and (0, 1), shifting (-1, 0) to (0, 0) makes it a RAT. 

Case 2 and Case 3:   No any possible shift results into a RAT.

Note that any point can be moved either in x- or in y- direction at a time by 1 unit.

Solution (in PYTHON):

Solution:
def dist(x1,y1,x2,y2):
	return (x1-x2)*(x1-x2)+(y1-y2)*(y1-y2)

def is_RAT(a):
	ab = dist(a[0],a[1],a[2],a[3])
	bc = dist(a[4],a[5],a[2],a[3])
	ca = dist(a[0],a[1],a[4],a[5])
	if ab == bc+ca or ca == bc+ab or bc == ab+ca:
		return True
	return False	
def samepoint(a):
	if a[0]==a[2] and a[1] == a[3]:
		return True
	if a[0]==a[4] and a[1] == a[5]:
		return True
	if a[4]==a[2] and a[5] == a[3]:
		return True
	return False	
t=input()
while t:
	t-=1
	a=[int(x) for x in raw_input().split()]
	d = [0,-1,+1]
	found = False
	for diff in d:
		for i in range(0,6):
			a[i]=a[i]+diff
			if is_RAT(a) and not samepoint(a):
				found = True
				break
			a[i]=a[i]-diff
		if found:
			break
	if found:
		print("YES")
	else:
		print("NO")